3.6 Summation convention and transformation of bases

Summation convention

In the above formulae summation over an index such as $i$ or $j$ invariably occurs on a pair of equal indices that are oppositely placed in the superscript and subscript position. Of course it is not inconceivable to have a summation between indices on the same leve but, as we shall see, it is unlikely to happen in a natural way. In fact, this phenomenon occurs with such regularity that it is possible to drop the summation sign $\sum_{i = 1}^{n}$ wheneve the same index $i$ appears in opposing positions without running into any serious misun derstandings, a convention first proposed by Albert Einstein (1879–1955) in the theory of general relativity, where multiple summation signs ofthis type arise repeatedly in the use of tensor calculus (see Chapter 18). The principal rule of Einstein’s summation convention is:

If, in any expression, a superscript is equal to a subscript then it will be assumed that these indice are summed over from 1 to $n$ where $n$ is the dimension of the space.

Repeated indices are called dummy or bound, while those appearing singly are called free. Free indices are assumed to take all values over their range, and we omit statements such as $(i, j = 1, 2, \dots, n)$ . For example, for any vector $u$ and basis ${e_{i}}$ it is acceptable to write

u = u^{i} e_{i} \equiv i = 1 \sum n u^{i} e_{i} .

The index $i$ is a dummy index, and can be replaced by any other letter having the same range,

u = u^{i} e_{i} = u^{j} e_{j} = u^{k} e_{k} .

For example, writing out Eqs. Eq. (3.6), Eq. (3.7) and Eq. (3.9) in this convention,

T e_{j} = T_{j}^{i} e_{i}, (3.11)

\tilde{v}^{i} = (T v)^{i} = T_{j}^{i} v^{j}, (3.12)

(S T)_{i}^{k} = S_{j}^{k} T_{i}^{j} . (3.13)

In more complicated expressions such as

T^{ij k} S_{hij} \equiv i = 1 \sum n j = 1 \sum n T^{ij k} S_{hij} (h, k = 1, \dots, n)

$i$ and $j$ are dummy indices and $h$ and $k$ are free. It is possible to replace the dummy indices

T^{i l k} S_{hi l} or T^{mik} S_{hmi}, etc.

without in any way changing the meaning ofthe expression. In such replacements any letter of the alphabet other than one already used as a free index in that expression can be used, but you should always stay within a specified alphabet such as Roman, Greek, upper case Roman, etc., and sometimes even within a particular range of letters.

Indices should not be repeated on the same level, and in particular no index should ever appear more than twice in any expression. This would occur in $V^{j} T_{ij}$ if the dummy index $j$ were replaced by the already occurring free index $i$ to give the non-permissible $V^{i} T_{ii}$ Although expressions such as $V_{j} T_{ij}$ should not occur, there can be exceptions; for example, in cartesian tensors all indices occur in the subscript position and the summation convention is often modified to apply to expressions such as $V_{j} T_{ij} \equiv \sum_{j = 1}^{n} V_{j} T_{ij}$

In an equation relating indexed expressions, a given free index should appear in the same position, either as a superscript or subscript, on each expression of the equation. For example, the following are examples of equations that are not permissible unless there are mitigating explanations:

T^{i} = S_{i}, T^{j} + U_{j} F^{j k} = S^{j}, T_{k}^{k k} = S^{k} .

A free index in an equation can be changed to any symbol not already used as a dummy index in any part of the equation. However, the change must be made simultaneously in al expressions appearing in the equation. For example, the equation

Y_{j} = T_{j}^{k} X_{k}

can be replaced by

Y_{i} = T_{i}^{j} X_{j}

without changing its meaning, as both equations are a shorthand for the $n$ equations

Y_{1} = T_{1}^{1} X_{1} + T_{1}^{2} X_{2} + \dots + T_{1}^{n} X_{n}

Y_{n} = T_{n}^{1} X_{1} + T_{n}^{2} X_{2} + \dots + T_{n}^{n} X_{n} .

Among the most useful identities in the summation convention are those concerning the Kronecker delta $δ_{j}^{i}$ defined by

δ_{j}^{i} = {10 if i = j, if i \neq = j . (3.14)

These are the components of the unit matrix, $I = [δ^{i}_{j}]$ . This is the matrix of the identity operator $id_{V}$ with respect to any basis ${e_{1}, \dots, e_{n}}$ . The Kronecker delta often acts as an ‘index replacement operator’; for example,

T_{k}^{ij} δ_{i}^{m} = T_{k}^{mj}, T_{k}^{ij} δ_{l}^{k} = T_{l}^{ij}, T_{k}^{ij} δ_{j}^{k} = T_{j}^{ij} = T_{k}^{ik} .

To understand these rules, consider the first equation. On the left-hand side the index $i$ is a dummy index, signifying summation from $i = 1$ to $i = n$ . Whenever $i \neq = m$ in this sum we have no contribution since $δ_{i}^{m} = 0$ , while the contribution from $i = m$ results in the right-hand side. The remaining equations are proved similarly.

Care should be taken with the expression $δ_{i}^{i}$ . If we momentarily suspend the summation convention, then obviously $δ_{i}^{i} = 1$ , but with the summation convention in operation the $i$ is a dummy index, so that

δ_{i}^{i} = δ_{1}^{1} + δ_{2}^{2} + \dots + δ_{n}^{n} = 1 + 1 + \dots + 1.

Hence

δ_{i}^{i} = n = dim V . (3.15)

In future, the summation convention will always be assumed to apply unless a rider like ‘summation convention suspended’ is imposed for some reason.

Basis transformations

Consider a change of basis

E = {e_{1}, e_{2}, \dots, e_{n}} ⟶ E^{'} = {e_{1}^{'}, e_{2}^{'}, \dots, e_{n}^{'}} .

By Theorem 3.5 each of the original basis vectors $e_{i}$ has a unique linear expansion in terms of the new basis,

e_{i} = A_{i}^{j} e_{j}^{'}, (3.16)

where $A_{i}^{j}$ represents the jth component of the vector $e_{i}$ with respect to the basis $E^{'}$ . Of course, the summation convention has now been adopted.

What happens to the components of a typical vector $v$ under such a change of basis? Substituting Eq. (3.16) into the component expansion of $v$ results in

v = v^{i} e_{i} = v^{i} A_{i}^{j} e_{j}^{'} = v^{' j} e_{j}^{'},

where

v^{' j} = A_{i}^{j} v^{i} . (3.17)

This law of transformation of components of a vector $v$ is sometimes called the contravariant transformation law of components, a curious and somewhat old-fashioned terminology that possibly defies common sense. Equation (3.17) should be thought of as a ‘passive’ transformation, since only the components of the vector change, not the physica vector itself. On the other hand, a linear transformation $S : V \to V$ of the vector space can be thought of as moving actual vectors around, and for this reason is referred to as an ‘active’ transformation.

Neverthless, it is still possible to think of Eq. (3.17) as a matrix equation if we represent the components $v^{i}$ and $v^{' j}$ of the vector $v$ as column vectors

v = v^{1} v^{2} ⋮ v^{n}, v^{'} = v^{'1} v^{'2} ⋮ v^{' n},

and the transformation coefficients $A_{i}^{j}$ as an $n \times n$ matrix

A = A_{1}^{1} A_{1}^{2} \dots A_{1}^{n} A_{2}^{1} A_{2}^{2} \dots A_{2}^{n} \dots \dots \dots \dots A_{n}^{1} A_{n}^{2} \dots A_{n}^{n} .

Equation (3.17) can then be written as a matrix equation

v^{'} = Av . (3.18)

Note, however, that $A = [A_{i}^{j}]$ is a matrix of coefficients representing the old basis ${e_{i}}$ in terms of the new basis ${e_{j}^{'}}$ . It is not the matrix of components of a linear operator.

Example 3.24

Let $V$ be a three-dimensional vector space with basis ${e_{1}, e_{2}, e_{3}}$ . Vectors belonging to $V$ can be set in correspondence with the $3 \times 1$ column vectors by
$v = v^{1} e_{1} + v^{2} e_{2} + v^{3} e_{3} ⟷ v = v^{1} v^{2} v^{3} .$
Let ${e_{i}^{'}}$ be a new basis defined by
$e_{1}^{'} = e_{1}, e_{2}^{'} = e_{1} + e_{2} - e_{3}, e_{3}^{'} = e_{1} + e_{2} + e_{3} .$
Solving for $e_{i}$ in terms of the $e_{j}^{'}$ gives
$e_{1} = e_{1}^{'}$ $e_{2} = - e_{1}^{'} + \frac{1}{2} (e_{2}^{'} + e_{3}^{'})$ $e_{3} = \frac{1}{2} (- e_{2}^{'} + e_{3}^{'}),$
and the components of the matrix $A = [A_{i}^{j}]$ can be read off using Eq. (3.16),
$A = 100 - 1 \frac{1}{2} \frac{1}{2} 0 - \frac{1}{2} \frac{1}{2} .$
A general vector $v$ is written in the $e_{i}^{'}$ basis as
$v = v^{1} e_{1} + v^{2} e_{2} + v^{3} e_{3} = v^{1} e_{1}^{'} + v^{2} (- e_{1}^{'} + \frac{1}{2} (e_{2}^{'} + e_{3}^{'})) + v^{3} \frac{1}{2} (- e_{2}^{'} + e_{3}^{'}) = (v^{1} - v^{2}) e_{1}^{'} + \frac{1}{2} (v^{2} - v^{3}) e_{2}^{'} + \frac{1}{2} (v^{2} + v^{3}) e_{3}^{'} = v^{'1} e_{1}^{'} + v^{'2} e_{2}^{'} + v^{'3} e_{3}^{'},$
where
$v^{'} \equiv v^{1} v^{2} v^{3} = v^{1} - v^{2} \frac{1}{2} (v^{2} - v^{3}) \frac{1}{2} (v^{2} + v^{3}) = Av .$
We will denote the inverse matrix to $A = [A^{i}_{j}]$ by $A^{'} = [A^{'}_{k}^{j}] = A^{- 1}$ . Using the sum mation convention, the inverse matrix relations
$A^{'} A = I, AA^{'} = I$
may be written componentwise as

$A_{j}^{' k} A_{i}^{j} = δ_{i}^{k}, A_{k}^{i} A_{j}^{' k} = δ_{j}^{i} . (3.19)$
From Eq. (3.16)
$A_{k}^{' i} e_{i} = A_{k}^{' i} A_{i}^{j} e_{j}^{'} = δ_{k}^{j} e_{j}^{'} = e_{k}^{'},$
which can be rewritten as

$e_{j}^{'} = A_{j}^{' k} e_{k} . (3.20)$

Exercise

From Eq. (3.19) or Eq. (3.20) derive the inverse transformation law of vector compo nents

$v^{i} = A_{j}^{' i} v^{j} . (3.21)$

We are now in a position to derive the transformation law of components of a linear operator $T : V \to V$ . The matrix components of $T$ with respect to the new basis, denoted $T^{'} = [T^{'}_{i}^{j}]$ , are given by

T e_{i}^{'} = T_{i}^{' j} e_{j}^{'},

and using Eqs. Eq. (3.16) and Eq. (3.20) we have

T e_{i}^{'} = T (A_{i}^{' k} e_{k}) = A_{i}^{' k} T e_{k} = A_{i}^{' k} T_{k}^{m} e_{m} = A_{i}^{' k} T_{k}^{m} A_{m}^{j} e_{j}^{'} .

Hence

T_{i}^{' j} = A_{m}^{j} T_{k}^{m} A_{i}^{' k}, (3.22)

or in matrix notation, since $A^{'} = A^{- 1}$

T^{'} = ATA^{'} = ATA^{- 1} . (3.23)

Equation (3.23) is the passive view – it represents the change in components of an operator under a change of basis. With a different interpretation Eq. (3.23) could however be viewed as an operator equation. If we treat the basis ${e_{1}, \dots, e_{n}}$ as fixed and regard $A$ as being the matrix representing an operator whose effect on vector components is given by

v^{' i} = A_{j}^{i} v^{j} ⟺ v^{'} = Av,

then Eq. (3.23) represents a change of operator, called a similarity transformation. If $x^{'} = A x, y^{'} = A y$ , then

y^{'} = A T x = A T A^{- 1} A x = T^{'} x^{'},

and $T^{'} = A T A^{- 1}$ is the operator that relates the transforms, under $A$ , of any pair of vectors $x$ and $y$ that were originally related through the operator $T$ . This is called the active view of Eq. (3.23). The two views are often confused in physics, mainly because operators are commonly identified with their matrices. The following example should help to clarify any lingering confusions.

Example 3.25

Consider a clockwise rotation of axes in $R^{2}$ through an angle $θ$ ,
$e_{1}^{'} = cos θ e_{1} - sin θ e_{2} e_{2}^{'} = sin θ e_{1} + cos θ e_{2} ⟺ e_{1} = cos θ e_{1}^{'} + sin θ e_{2}^{'} e_{2} = - sin θ e_{1}^{'} + cos θ e_{2}^{'} .$
The matrix of this basis transformation is
$A = [A_{j}^{i}] = (cos θ sin θ - sin θ cos θ)$
and the components of any position vector
$x = (y x)$
(b)

Figure 3.1 Active and passive views of a transformation

Figure 3.2 Active view of a similarity transformation

change by
$x^{'} = Ax ⟺ x^{'} = cos θ x - sin θ y y^{'} = sin θ x + cos θ y .$
This is the passive view. On the other hand, if we regard $A$ as the matrix of components of an operator with respect to fixed axes $e_{1}, e_{2}$ , then it represents a physical rotation of the space by an angle $θ$ in a counterclockwise direction, opposite to the rotation of the axes in the passive view. Figure 3.1 demonstrates the apparent equivalence of these two views, while Fig. 3.2 illustrates the active view of a similarity transformation $T^{'} = A T A^{- 1}$ on a linear operator $T : R^{2} R^{2}$

Problems

Problem 3.11

Let ${e_{1}, e_{2}, e_{3}}$ be a basis ofa three-dimensional vector space $V$ . Show that the vectors ${e_{1}^{'}, e_{2}^{'}, e_{3}^{'}}$ defined by
$e_{1}^{'} = e_{1} + e_{3} e_{2}^{'} = 2 e_{1} + e_{2} e_{3}^{'} = 3 e_{2} + e_{3}$
also form a basis of $V .$

What are the elements of the matrix $A = [A_{i}^{j}]$ in Eq. (3.16)? Calculate the components of the vector
$v = e_{1} - e_{2} + e_{3}$
with respect to the basis ${e_{1}^{'}, e_{2}^{'}, e_{3}^{'}}$ , and verify the column vector transformation $v^{'} = A v$

Problem 3.12

Let $T : V \to W$ be a linear map between vector spaces $V$ and $W$ . $If {e_{i} ∣ i = 1, \dots, n}$ is a basis of $V$ and ${f_{a} ∣ a = 1, \dots, m}$ a basis of $W_{i}$ , how does the matrix $T$ , defined in Problem 3.9, transform under a transformation of bases
$e_{i} = A_{i}^{j} e_{j}^{'}, f_{a} = B_{a}^{b} f_{b}^{'} ?$
Express your answer both in component and in matrix notation.

Problem 3.13

Let $e_{1}, e_{2}, e_{3}$ be a basis for a three-dimensional vector space and $e_{1}^{'}, e_{2}^{'}, e_{3}^{'}$ a second basis given by
$e_{1}^{'} = e_{3}, e_{2}^{'} = e_{2} + 2 e_{3}, e_{3}^{'} = e_{1} + 2 e_{2} + 3 e_{3} .$
(a) Express the $e_{i}$ in terms of the $e_{j}^{'}$ , and write out the transformation matrices $A = [A^{i}_{j}]$ and $A^{'} = A^{- 1} = [A^{' i}_{j}] .$

(b) If $u = e_{1} + e_{2} + e_{3}$ , compute its components in the $e_{i}^{'}$ basis.

(c) Let $T$ be the linear transformation defined by
$T e_{1} = e_{2} + e_{3}, T e_{2} = e_{3} + e_{1}, T e_{3} = e_{1} + e_{2} .$
What is the matrix of components of $T$ with respect to the basis $e_{i} ?$

(d) By evaluating $T e_{1 :}^{'}$ , etc. in terms of the $e_{j}^{'}$ , write out the matrix of components of $T$ with respect to the $e_{j}^{'}$ basis and verify the similarity transformation $T^{'} = AT A^{- 1}$

Szekeres · 数学物理

章节

3.6 Summation convention and transformation of bases

Summation convention

Basis transformations

Problems

关系图谱

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